Polynomial maps with nilpotent Jacobians in dimension three II
Abstract
In the paper, we first classify all polynomial maps of the form in the case that is nilpotent and , . Then we classify all polynomial maps of the form in the case that is nilpotent and , . Finally, we classify polynomial maps of the form in certain conditions.
Keywords. Jacobian Conjecture, Nilpotent Jacobian matrix, Polynomial maps
MSC(2010). Primary 14E05; Secondary 14A05;14R15
1 Introduction
Throughout this paper, we will write for algebraically closed field and ()for the polynomial algebra over with () indeterminates. Let be a polynomial map, that is, for all . Let be the Jacobian matrix of .
The Jacobian Conjecture (JC) raised by O.H. Keller in 1939 in [8] states that a polynomial map is invertible if the Jacobian determinant is a nonzero constant. This conjecture has been attacked by many people from various research fields, but it is still open, even for . Only the case is obvious. For more information about the wonderful 70-year history, see [1], [5], and the references therein.
In 1980, S.S.S.Wang ([9]) showed that the JC holds for all polynomial maps of degree 2 in all dimensions (up to an affine transformation). The most powerful result is the reduction to degree 3, due to H.Bass, E.Connell and D.Wright ([1]) in 1982 and A.Yagzhev ([11]) in 1980, which asserts that the JC is true if the JC holds for all polynomial maps , where is homogeneous of degree 3. Thus, many authors study these maps and led to pose the following problem.
(Homogeneous) dependence problem. Let be a (homogeneous) polynomial map of degree such that is nilpotent and . Whether are linearly dependent over ?
The answer to the above problem is affirmative if rank ([1]). In particular, this implies that the dependence problem has an affirmative answer in the case . D. Wright give an affirmative answer when is homogeneous of degree 3 in the case ([10]) and the case is solved by Hubbers in [7]. M. de Bondt and A. van den Essen give an affirmative answer to the above problem in the case is homogeneous and ([3]). A. van den Essen finds the first counterexample in dimension three for the dependence problem ([6]). M. de Bondt give a negative answer to the homogeneous dependence problem for . In particular, he constructed counterexamples to the problem for all dimensions ([2]). In [4], M. Chamberland and A. van den Essen classify all polynomial maps of the form with nilpotent. In particular, they show that all maps of this form with , nilpotent and are linearly independent has the same form as the counterexample that gave by A. van den Essen in [6] (up to a linear coordinate change). We classify all polynomial maps of the form in the case that is nilpotent and , ([12]) and classify all polynomial maps of the form in the case that is nilpotent and or or is a prime number ([13]).
In section 2, we classify all polynomial maps of the form in the case that is nilpotent and , . Then, in section 3, we classify all polynomial maps of the form in the case that is nilpotent and , . We prove that are linearly dependent in the case that is nilpotent and has the form: with and , in section 4. The main results in the paper are Theorem 2.2, Theorem 2.5, Theorem 3.2 and Theorem 3.3, Theorem 4.1. We define and that is the highest degree of in .
2 Polynomial maps of the form
In this section, we classify all polynomial maps of the form in the case that is nilpotent and , .
Lemma 2.1.
Let and with . If or for some , then is a polynomial of for some .
Proof.
Let . Then we have and . Thus, we have . We always view that the polynomials are in with coefficients in when comparing the coefficients of .
Case I If , then we have
for some . That is,
Comparing the coefficients of of the above equation, we have that . Then let , , it follows from equation that . That is, . Let . Then the conclusion follows.
Case II If , then . That is,
Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (2.2)
Comparing the coefficients of and of equation
, we have that and
. Then equation has the
following form:
Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (2.3)
Claim:
for and .
Comparing the coefficients of of equation , we have
the following equation:
Suppose . Then comparing the coefficients of of equation , we have the following equation:
That is,
Thus, we have that for . Then equation has the following form: . That is,
Thus, we have . Therefore, we have . Let . Then the conclusion follows. β
Theorem 2.2.
Let be a polynomial map over . Assume that and the components of are linearly independent over . If is nilpotent and , then , , , where ; ; , and , .
Proof.
If , then the conclusion follows from Theorem 3.5 in [13]. Let . Then it follows from Lemma 3.2 in [13] that and . Since is nilpotent, so we have the following equations:
It follows from equation that . Thus, there exists such that
It follows from equation and Lemma 3.1 in [13] that there exists such that
Since and , so we have or 3.
If , then the conclusion follows from the proof of Theorem 2.8 in [12].
If , then . Let
.
Claim: .
Let . It
follows from Lemma 3.4 in [13] that . It
follows from equations and that
. Thus, we have
. Substituting equations and to
equation , we have the following equation:
(1) If is irreducible, then we have or . Since is a polynomial of , so we have with and , . Since is irreducible, so there exists such that in the case that . It follows from Lemma 2.1 that is a polynomial of for some . That is, . Then the conclusion follows from the proof of Theorem 2.8 in [12].
(2) If is reducible, then
for some . Clearly, , are
irreducible. It follows from equation that we have the
following cases:
Case I
Case II
Case III and
Case IV in the case .
Case V in the case .
Case I If , then the conclusion follows from Lemma 2.1.
Case II If , then . Since is a polynomial of , so for and , . Thus, there exist such that and . That is,
and
We always view that the polynomials are in with coefficients in in the following arguments. Comparing the coefficients of , of equations and , we have the following equations:
and
It follows from equations and that
and
Then equations and have the following forms:
and
Then equation has the following form:
Thus, we have . That is, for
some . Therefore, we have
. Then
for some
. Since , so we have
. That is, . Thus, is a
polynomial of . Therefore, . Then
the conclusion follows from the proof of Theorem 2.8 in [12]
Case III Since for and , . Clearly, is irreducible, so there exists such that . That is,
and
where . Comparing the coefficients of , of equation , we have the following equations:
and
Comparing the coefficients of , of equation , we have the following equations:
and
Substituting equations and to equation , we have the following equation:
Integrating the two sides of the above equation with respect to , we have the following equation:
for some . Comparing the coefficients of , and of equation , we have the following equations:
It follows from equations and that
Substituting equations , and to equation , we have the following equation:
Comparing equation with equation , we have . That is,
Thus, we
have . That is, for
some . It follows from equation
that . Then we have
for some . Since , so
we have . Thus, we have
. Therefore, is a polynomial of .
That is, . Then the conclusion follows
from the proof of Theorem 2.8 in [12].
Case IV, V If , then it follows from equation that . Thus, we have for some .
If , then . Thus, we have . That is, . Therefore, we have .
If , then it follows from the arguments of Case II
that for some . That is,
. Thus, we have
. Therefore, .
Thus, are polynomials of in the two cases. Then the
conclusion follows from the proof of Theorem 2.8 in [12].
β
Remark 2.3.
We can replace the condition that by the condition in Theorem 2.10 and replace the condition that by the condition in Theorem 3.2 and Theorem 3.4 in [12].
Corollary 2.4.
Let be a polynomial map over . Assume that and the components of are linearly independent over . If is nilpotent and or or or , then has the form of Theorem 2.2.
Proof.
Let . Then it follows from Lemma 3.2 in [13] that and . Since is nilpotent, so we have the following equations:
It follows from equation and Lemma 3.1 in [13] that there exists such that
Thus, we have .
If or , then the conclusion follows from Corollary 3.9 in [13].
If or or or , then the conclusion follows from Theorem 3.3 in [13].
If or 9 or or 9, then or 2 or 3 or 6 or 9.
Case I If or 2 or 3, then the conclusion follows from the proof Theorem 2.2.
Case II If , then or .
(1) If , then it follows from equation that with , . That is, . Thus, is a polynomial of . Then the conclusion follows from Theorem 2.1 in [4].
(2) If , then it follows from the arguments of (1) that is a polynomial of . It follows from Corollary 2.3 in [13] that are linearly dependent. This is a contradiction!
Case III If , then or . That is, or . Then the conclusion follows from the arguments of Case II. β
Theorem 2.5.
Let be a polynomial map over . Assume that and the components of are linearly independent over . If is nilpotent and the variety is irreducible for any and , then has the form of Theorem 2.2.
Proof.
Let . Then it follows from Lemma 3.2 in [13] that and . Since is nilpotent, so we have the following equations:
It follows from equation that . Thus, there exists such that
It follows from equation and Lemma 3.1 in [13] that there exists such that
If , then the conclusion follows from the proof of
Theorem 2.8 in [12].
Suppose . Let with . It follows from equation and Lemma 3.4 in [13] that
It follows from equations , and that
Since , , so we have and . Thus, the variety is irreducible. That is, there exists an irreducible polynomial such that
for some . It follows from equation that
. That is, . Since
is irreducible, so we have or
.
Case I Suppose . Since is a polynomial of , so it follows from the Fundamental Theorem of Algebra that for and , . If , then there exists such that . That is,
for some . Thus, it follows from equations and that . That is,
Since is irreducible and , so it follows from the above equation that . That is,
for some . Thus, it follows from equations and that
Then it follows from equation that . That is,
Since is irreducible and , so we have . We can do step by step until
for some . Then it follows from the above equation and equation that . That is,
Let ,
with and . It follows from
equations and that .
That is, . It follows from equation
that
Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (2.37)
We always view that the polynomials are in with
coefficients in when comparing the coefficients of
.
Suppose . Comparing the coefficients of of equation , we have the following equation:
Since , so we have . This is a contradiction because and . Therefore, we have . That is, . Then equation has the following form:
Thus, we have . That is, with
. So is a polynomial of with
. That is, . Then
the conclusion follows from the proof of Theorem 2.8 in [12].
Case II If , then
for some . Thus, it follows from equations and that , . Thus, we have the following equation:
Since is irreducible and , so we have . That is, for some . It follows from equation that . We can do step by step until . Then . That is,
Since , and , so we have . Thus, we have . That is, . Then equation has the following form:
where . Let , . Then it follows from equation that . That is, , where . Thus, are polynomials of . Then the conclusion follows from the proof of Theorem 2.8 in [12]. β
Corollary 2.6.
Let be a polynomial map over . Assume that and the components of are linearly independent over . If is nilpotent and the variety or the variety is irreducible, then has the form of Theorem 2.2.
Proof.
Let . Then it follows from Lemma 3.2 in [13] that and . Since is nilpotent, so we have the following equations:
It follows from equation and Lemma 3.1 in [13] that there exists such that
Since or is irreducible and is a subvariety, so is irreducible. Then the conclusion follows from the proof of Theorem 2.5. β
3 Polynomial maps of the form
In this section, we classify polynomial maps of the form in the case that is nilpotent and . Combining Theorem 2.2 with Corollary 3.7 in [13], we have the following Proposition.
Proposition 3.1.
Let be a polynomial map over . Assume that and the components of are linearly independent over . If is nilpotent and or at least one of , is a prime, then , , , where , , , and , .
In the following theorem, we denote , and .
Theorem 3.2.
Let be a polynomial map over . Assume that and . If is nilpotent, then are linearly dependent.
Proof.
If , then it follows from Proposition 2.1 in [12] that are linearly dependent. Thus, we can assume that . Since is nilpotent, we have the following equations:
Let ,
with and , , for , . It follows from
equation that
Β Β Β Β Β Β Β Β Β Β Β Β Β Β (3.4)
We always view that the polynomials are in with
coefficients in in the following arguments.
Since , comparing the coefficients of of equation , we have . That is,
Suppose . Then . Thus, we have
by integrating the two sides of the above
equation with respect to , where is a function of .
Since and is a function of .
Thus, we have . This is a contradiction!
Therefore, we have . Then we have by
comparing the coefficients of of equation for
. Then equation has the following
form:
Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (3.5)
Comparing the coefficients of for of
equation , we have the following equations:
Then we have
by comparing the coefficients of of equation for . If , then for . If , then we have by comparing the degree of of two sides of equation for . Since , we have . This is a contradiction! Thus, we have and . Therefore, we have . Then equation has the following form:
Since , so it follows from equation (3.3) that . That is,
Then we have by comparing the coefficients of
of equation . Thus, we have or .
(1) If , then it follows from equation that or .
(i) If , then because . Thus, are linearly dependent.
(ii) If , then . It follows from equation that . That is, Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (3.10)
If , then it follows from equation that . Since , we have . Thus, are linearly dependent.
If , then we have by
comparing the coefficients of
of equation respectively.
If , then we have by comparing the coefficient of
of equation . Since , we have .
Thus, are linearly dependent.
If , then it follows from equation () that . This is a contradiction!
(2) If , then . It follows from equation
that
Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (3.11)
If , then we have by comparing the coefficient of of equation . Thus, we have by comparing the coefficients of of equation for . Comparing the coefficients of of equation , we have . Thus, we have because . So are linearly dependent.
If , then equation has the following form:
Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (3.12)
Thus, we have
by comparing the coefficients of of equation for . Then equation has the following form:
Comparing the coefficients of of equation for , we have
for . Then equation has the following form:
It follows from equation that
for some . Substituting equation (3.17) to equation (3.13) for , we have
Thus, we have by integrating the two sides of the above equation with respect to , where . That is,
Substituting equations and to equation for , we have
If , then it follows from equation that .
Then it reduces to (1).
If , then we have
Thus, we have and
by comparing the degree of of the equation (3.19). Then we have
by substituting equation to equation . Substituting equation (3.20) to equation (3.21), we have
If , then it follows from equation (3.20) that . Thus, it follows from equation (3.18) that . Since , we have . Therefore, are linearly dependent.
If , then equation (3.22) has the following form:
Substituting equations (3.18) and (3.23) to equation (3.14), we have the following equation:
Substituting equation (3.8) to the above equation, we have . It follows from equation (3.20) that . This is a contradiction! β
Theorem 3.3.
Let be a polynomial map over . Assume that and the components of are linearly independent over . If is nilpotent and , then there exists such that has the form of Proposition 3.1.
Proof.
It follows from Theorem 3.2 that . If or , then it reduces to Proposition 2.1 in [12] and Proposition 3.1 respectively. Thus, we can assume that and . Since is nilpotent, we have the following equations:
Let and with .
It follows from equation (3.1) that
Β Β Β Β Β Β Β Β Β Β Β Β Β Β (3.24)
Then we have
and
by comparing the coefficients of and of equation (3.24) respectively.
It follows from equation (3.2) that
Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (3.27)
Comparing the coefficients of of equation (3.27) for
, we have
Then we have
and
by comparing the coefficients of and of equation (3.27) respectively.
It follows from equation (3.3) that
Then we have
and
by comparing the coefficients of and of equation (3.31)
orderly.
If , then let . Thus, we have . It follows from equations (3.32) and (3.33) that
and
It follows from equations (3.28) and (3.34) that
If , then it follows from equation (3.32) that because . That is, .
If , then we have
for . Substituting equations (3.34), (3.35), (3.36) to equation (3.29), we have the following equation:
Substituting equations (3.34), (3.35), (3.36), (3.37) to equation (3.30), we have the following equation:
That is,
Substituting equation (3.34) to the above equation, we have
Then we have by comparing the degree of of two
sides of the above equation. This is a contradiction!
Consequently, we have with . Let
Then
and , . Since , so the conclusion
follows from Proposition 3.1.
If , then it follows from equation (3.32) that or .
(1) If , then because . Thus, are linearly dependent. This is a contradiction!
(2) If , then it follows from equation that or .
Case I If , then . Thus, the conclusion follows from the former arguments.
Case II If , then it follows from equations (3.25) and (3.26) that . Thus, it follows from equation (3.29) that or . If , then we have , it reduces to Case I. If , then it follows from equation (3.30) that . Thus, we have because . Therefore, are linearly dependent. This is a contradiction! β
Corollary 3.4.
Let be a polynomial map with . Assume that the components of are linearly independent over . If is nilpotent and , then there exists such that has the form of Proposition 3.1.
Proof.
Let
Then . Since is nilpotent, we have that is nilpotent. Since , and , so it follows from Theorem 3.3 that there exists such that is of the form of Proposition 3.1. Let . Then the conclusion follows. β
4 Polynomial maps of the form
In the section, we classify polynomial maps of the form in the case that is nilpotent and . Firstly, we prove that are linearly dependent in the case that is nilpotent and and . In the proof of the following theorem, we divide two cases according to the degree of in . Case I: ; Case II: . In the first case, we divide two parts. We have three subcases in the second case.
Theorem 4.1.
Let be a polynomial map with . Assume that and . If is nilpotent, then are linearly dependent.
Proof.
Let , with and . Since is nilpotent, we have the following equations:
It follows from equation (4.1) that
We always view that the polynomials are in with coefficients in in the following arguments. Thus, we have
and
for by comparing the coefficients of of equation for .
It follows from equations (4.2) and (4.5) that
Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (4.7)
Case I If , then comparing the coefficients of and of equation (4.7), we have and
Thus, we have or .
(1) If , then it follows from equations (4.5) and (4.8) that and
It follows from equation (4.6) that , so there exists such that , . It follows from equation (4.9) that . Then it follows from Lemma 2.1 in [4] that
and
for some and , , . Then equation (4.7) has the following form:
Thus, we have the following equations
and
by comparing the coefficients of and of equation (4.12). It follows from equations (4.3) and (4.5) that
Comparing the coefficients of of equation , we have the following equations:
and
and
It follows from equations (4.11) and (4.16) that . Thus, we have or or .
(i) If , then
It follows from equation (4.13) that . That is,
Integrating with respect to of two sides of the above equation, we have
for some . Substituting equations (4.19) and (4.20) to equation (4.17), we have the following equation:
Thus, we have or and .
If , then it follows from equations (4.13) and (4.19) that . It follows from equation (4.18) that . Thus, we have or . If , then it follows from equation (4.14) that . If , then because . Thus, are linearly dependent.
If , then and . Let and . Then it follows from equation (4.19) that and for some . Since , so it follows from equation (4.20) that
It follows from equations (4.14) and (4.21) that
It follows from equations (4.18) and (4.21) that
Substituting equation (4.22) to equation (4.23), we have the following equation:
Since , so the above equation has the following form:
Substituting equation (4.6) to the above equation, we have . That is, . Since , so we have . It follows from equation (4.21) that . Thus, are linearly dependent.
(ii) If , then because . That is, ,
. It follows from equation (4.13) that
. It follows from equation (4.17) that . Thus,
we have or
.
If , then because . Thus, are
linearly dependent.
If , then it follows from equation (4.18) that
. That is, or . If , then
it follows from equation (4.14) that . Thus, we have that
. It reduces to the above case.
(iii) If , then we have
because we can assume that . Otherwise, it reduces to the former two cases. Let , . Then we have that . That is, . It follows from equations (4.10), (4.11), (4.17),(4.24) that
It follows from equations (4.18) and (4.24) that
It follows from equations (4.25) and (4.26) that or
and .
If , then it follows from equation (4.24) that . Thus,
we have because . so are linearly
dependent.
If and , then . Thus, it follows from equations (4.10), (4.11) and
(4.13) that . It reduces to the former case.
(2) If , then it follows from equation (4.8) that
It follows from equation (4.7) that
Comparing the coefficients of of equation (4.28), we have the following equations:
It follows from equations (4.3) and (4.5) that
Comparing the coefficients of and of the above equation, we have the following equations:
It follows from equation (4.32) that or .
If , then it follows from equation (4.33) that . Thus, we have that or . If , then because . Thus, are linearly dependent. If , then it follows from equation (4.6) that
Substituting equations (4.35) and (4.6) to equation (4.29), we have
the following equation: . Thus, we
have or .
If , then it follows from equation (4.29) that
. Thus, we have or . If
, then it reduces to Case I(1). If , then it
follows form equation (4.34) that . Thus, we have
or . If , then it follows from equation
(4.31) that . Therefore, we have because . So
are linearly dependent.
If , then we have
Suppose . Then we have
by integrating of two sides of equation (4.36) with respect to . where is a function of . Since , we have . That is, for and . Then it follows from equations (4.29) and (4.6) that
That is,
If , then we have that by comparing the
degree of of equation (4.37). Thus, we have . This is
a contradiction. Therefore, we have . It follows
from equation (4.36) that , This is a contradiction! If
, then it follows from equation (4.36) that
. which reduces to case I(1).
Case II If , then we have
and
by comparing the coefficients of and of equation
(4.7). It follows from equations (4.38) and (4.39) that
or or .
(a) If , then equation (4.7) has the following form:
If , then by comparing the coefficient of of
equation (4.40). Thus, it follows from equation (4.3) that
. Therefore, we have or
. If , then it follows from
equation (4.40) that . Thus, we have because
. Therefore, are linearly dependent.
If , then comparing the coefficients of and of
equation (4.40), we have the following equations:
It follows from equations (4.3) and (4.5) that
Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (4.44)
Comparing the coefficients of of the above equation, we have
. Thus, we have or
.
(1) If , then it follows from equation (4.44) that . That is, or . If , then it follows from equation (4.40) that . Thus, we have because . Therefore, are linearly dependent.
(2) If , then
If , then . It follows from equation (4.41) that . Then it follows from the arguments of Case II (1) that are linearly dependent. We can assume that . Then we have that by integrating the two sides of equation (4.45) with respect to , where is a function of . Since , we have . That is,
with , . Substituting equations (4.6) and to equation (4.41), we have that
Then we have by comparing the degree of of equation
(4.47). It follows from equation (4.47) that . Then it
follows from the arguments of Case II(1) that are linearly
dependent.
(b) If , then it follows from equation (4.6) that
. then equation (4.7) has the following form:
Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (4.48)
Then we have the following equations:
by comparing the coefficients of of equation (4.49) for . Comparing the coefficients of and of equation (4.48), we have the following equations:
It follows from equations (4.3) and (4.5) that
Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (4.53)
Then we have by comparing the coefficients of of
equation (4.53). That is, or .
(3) If , then equation (4.53) has the following form :
Comparing the coefficients of of equation (4.54), we have that . That is, or . If , then it reduces to Case II(a). If , then because . Thus, are linearly dependent.
(4) If , then comparing the coefficients of and of equation (4.53), we have
and
If , then it follows from equation (4.49) that
If , then it follows from equation (4.50) that
Substituting equation (4.57) to equation (4.55), we have
for . Thus, we have or
If , then it reduces to Case II(a). Otherwise, substituting equations (4.57) and (4.58) to equation (4.52),we have that
Substituting equations (4.57) and (4.58) to equation (4.56), we have that
Then we have by substituting
equations (4.6) and (4.59) to equation (4.60). That is,
. It follows from equation (4.49) that
. Then it reduces to Case II (3).
(c) If , then it follows from equations (4.2)
and (4.5) that
Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (4.61)
Comparing the coefficients of of equation (4.61) for
, we have the following equations:
Comparing the coefficients of and , we have the following equations:
It follows from equations (4.3) and (4.5) that
Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (4.67)
Comparing the coefficients of and of equation
(4.67), we have the following equations:
That is, and . Then equation (4.67) has the following form :
Then we have by comparing the coefficients of of equation (4.68) for . Thus, we have or . If , then it follows from equation (4.63) that . If , then it follows from equation (4.68) that or . If , then it follows from equation (4.63) and (4.62) that or . If , then it reduces to Case II(b). If , then it follows from equation (4.64) that . It follows from equation (4.65) that . Then it follows from equation (4.66) that . Thus, we have because . Therefore, are linearly dependent. β
Corollary 4.2.
Let be a polynomial map with
. Assume that and the components of
are linearly independent over . If is nilpotent,
then
has the form of Theorem 2.2 in the case and
.
has the form of Proposition 3.1 in the case
and .
there exists such that
has the form of Theorem 2.2 in the case and for some .
Proof.
Since are linearly independent, so it follows from Theorem 4.1 that . Thus, we have and . Then the conclusions (1) and (2) follow from Theorem 2.2 and Proposition 3.1 respectively. The conclusion of (3) follows from the proof of Theorem 3.10 in [12] and Theorem 2.2. β
References
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